a) Vì \(\left|2x+4\right|\ge0;\left|y\right|\ge0\)

mà \(\left|2x+4\right|+\left|y\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|2x+4\right|=0\\\left|y\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-2;0\right)\)

dạ cho mình hỏi bạn biết làm câu b không ạ

a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)

\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)

mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)

\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

b) Tương tự câu a, ta có:

\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

c. Tương tự, ta có:

\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...

b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...

c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...

\(\frac{x}{5}=\frac{y}{3}\)và x²-y²=4(x,y>0)

\(\Rightarrow\frac{x}{5}=\frac{y}{3}=\frac{x^2}{5^2}=\frac{y^2}{3^2}=\frac{x^2-y^2}{25-9}=\frac{4}{16}=\frac{1}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\Rightarrow\frac{x^2}{25}=\frac{1}{4}\Rightarrow x^2=\frac{25}{4}\Rightarrow x=\frac{5}{2}\)

\(\Rightarrow\frac{y^2}{9}=\frac{1}{4}\Rightarrow y^2=\frac{9}{4}\Rightarrow y=\frac{3}{2}\)

Vậy x =\(\frac{5}{2}\)và y =\(\frac{3}{2}\)

Ta có:

\(\frac{x}{3}=\frac{y}{5}\Rightarrow\frac{x^2}{3}=\frac{y^2}{5}\)

Áp dụng dãy tỉ số bằng nhau, ta có:

\(\frac{x^2}{3^2}=\frac{y^2}{5^2}=\frac{x^2-y^2}{3^2-5^2}=\frac{-4}{-16}=\frac{1}{4}\)

\(\Rightarrow\frac{x^2}{3^2}=\frac{1}{4}\Rightarrow x=\sqrt{3^2.\frac{1}{4}}=\frac{3}{2}\)

\(\frac{y^2}{5^2}=\frac{1}{4}\Rightarrow y=\sqrt{5^2.\frac{1}{4}}=\frac{5}{2}\)

a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{1}{2}\)

c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)

a, Xét : x-4 = 0 => x= 4

            2x+1 = 0 => x= \(\frac{1}{2}\)

            x+3 = 0 => x = -3

            x + 9 = 0 => x = -9

Khi đó ta có bảng xét dấu : 

=> có 5 trường hợp:

TH1 : \(x\le-9\)

TH2 : \(-9\le x< -3\)

TH3 : \(-3\le x< \frac{1}{2}\)

TH4 : \(\frac{1}{2}\le x< 4\)

Do đó :

TH1 : \(x\le-9\)

Ta có :  /x-4/ = -(x-4) = 4 - x

            /2x+1/ = -(2x+1) = -2x -1

           /x+3/   = -(x + 3 ) = -x - 3

          /x-9/ = -(x-9) = -x + 9                  Thay vào đề bài ta có:

                                               3.(4-x) + 2x-1 +5(-x - 3) -x-9 = 5

                                    => 12 - 3x + 2x - 1 + -5x - 15 - x - 9 = 5

                                    =>(12 - 1 - 15 -9 ) +(-3x +2x -5x -x) = 5

                                   => -13 - 7x                                        = 5

                                             7x                                =     -13 - 5

                                                 7x =      -18

                                              x = \(\frac{-18}{7}\)( Ko TM)

Tương tự với 4 trường hợp còn lại.

\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y-z=26\)

\(BCNN\left(3,5\right)=15\)

\(\Rightarrow\frac{x}{2}=\frac{y}{3}\)\(\Rightarrow\frac{x}{10}=\frac{y}{15}\)(1)

      \(\frac{y}{5}=\frac{z}{4}\)\(\Rightarrow\frac{y}{15}=\frac{z}{12}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y-z}{10+15-12}=\frac{26}{13}=2\)

\(\Rightarrow x=2.10=20\)

     \(y=2.15=30\)

     \(z=2.12=24\)

Vậy x = 20 ; y = 30 ; z = 24

Đáp án là:

a) x=2,5;y=1,5.

Đáp án là:

a) x=2,5;y=1,5.

b) Hầu như x=29,46230884.